3.85P-2.00t + 3.00, where x is in meters and t is in An object moves along the x

Question

3.85P-2.00t + 3.00, where x is in meters and t is in An object moves along the x axis according to the equation x seconds (a) Determine the average speed between t- 2.80 s and t-4.80 s (b) Determine the instantaneous speed at t- 2.80 s. Determine the instantaneous speed att 4.80 s. (c) Determine the average acceleration between t2.80 s and t 4.80 s. m/s2 (d) Determine the instantaneous acceleration at t 2.80 s. m/s2 Determine the instantaneous acceleration at t -4.80 s. m/s2 (e) At what time is the object at rest?

Explanation / Answer

given

x= 3.85 t2 - 2t + 3  

a) we know that

at t = 2.8 ====> xi = 27.584 m

at t= 4.8 s ===> xf = 82.104 m

so avg speed = distance tarvelled / time taken = xf-xi / t = (82.104-27.584)/(4.8-2.8) = 27.26 m/s

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b) we know that

v = dx/dt = d(3.85 t2 - 2t + 3 )/dt = (7.7 t - 2 )

at t= 2.80 s ===> v1 = 7.7x2.80 - 2 = 19.56 m/s

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c)

at 4.8 s =====> v2 = 7.7 x4.8 - 2 = 34.96 m/s

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d)

here time taken = 4.8-2.8 = 2 sec

so avg acc = (v2-v1)/t = (34.96 - 19.56) / 2 = 7.7 m/s2

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e) we know that

a = dv/dt = d(7.7 t - 2 )/dt = 7.7 m/s2

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f) t=? when v=0

so put (7.7 t - 2 ) = 0

t = 2/7.7 = 0.259 sec

answer

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